Two friends agree to meet at a park with the following conditions. Each will reach the park between 4:00 pm and 5:00 pm and will see if the other has already arrived. If not, they will wait for 10 minutes or the end of the hour whichever is earlier and leave. What is the probability that the two will not meet?

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3 Answers

+19 votes

Best answer

We are given that both will be reaching the park between $4:00$ and $5:00$.

Probability that one friend arrives between $4:00$ and $4:50 = \dfrac{5}{6}$
Probability that one friend arrives between $4:00$ and $4:50$ and meets the other arriving in the next $10$ minutes $=\dfrac{5}{6}\times \dfrac{1}{6}\times {2} =\dfrac{10}{36} =\dfrac{5}{18}.$
(For any time of arrival between $4:00$ and $4:50,$ we have a $10$ minute interval possible for the second friend to arrive, and $2$ cases as for choosing which friend arrives first)

Probability that both friend arrives between $4:50$ and $5:00 =\dfrac{1}{6}\times \dfrac{1}{6} =\dfrac{1}{36}.$

This covers all possibility of a meet. So, required probability of non-meet

@Rajesh Pradhan This case is included when he says the following:
The probability that one friend arrives between 4:00 and 04:50 and meets the other arriving in the next 1010 minutes =56×16×2=1036=518.=56×16×2=1036=518.
(For any time of arrival between 4:00 and 4:50, we have a 10 minute interval possible for the second friend to arrive, and 2 cases as for choosing which friend arrives first)

@Prateek , sorry , I am not asking about it. It is already mentioned in Arjun sir's answer. I am asking about calculation part means how your calculation part of area related to Arjun sir's calculation part ?

I thought that both approaches are inter-related and Arjun sir is also finding the required area using his calculation indirectly. But Now, I think both are different approaches. sorry bro I have wasted your time :(